Difference between revisions of "2017 AMC 12A Problems/Problem 5"
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<math>|A||B|+{|B|\choose 2} = 20\cdot 10+{10\choose 2} = 200+45 = \boxed{(B)=\ 245}</math> | <math>|A||B|+{|B|\choose 2} = 20\cdot 10+{10\choose 2} = 200+45 = \boxed{(B)=\ 245}</math> | ||
| − | == | + | ==Solution - Complementary Counting== |
The number of handshakes will be equivalent to the difference between the number of total interactions and the number of hugs, which are <math>{30\choose 2}</math> and <math>{20\choose 2}</math>, respectively. Thus, the total amount of handshakes is <math>{30\choose 2} - {20\choose 2} = 435 - 190= \boxed{(B)=\ 245} </math> | The number of handshakes will be equivalent to the difference between the number of total interactions and the number of hugs, which are <math>{30\choose 2}</math> and <math>{20\choose 2}</math>, respectively. Thus, the total amount of handshakes is <math>{30\choose 2} - {20\choose 2} = 435 - 190= \boxed{(B)=\ 245} </math> | ||
Revision as of 16:22, 18 February 2017
Problem
At a gathering of 
people, there are 
people who all know each other and 
people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?
Solution
Let the group of people who all know each other be 
, and let the group of people who know no one be 
. Handshakes occur between each pair 
such that 
and 
, and between each pair of members in . Thus, the answer is
Solution - Complementary Counting
The number of handshakes will be equivalent to the difference between the number of total interactions and the number of hugs, which are 
and 
, respectively. Thus, the total amount of handshakes is 
See Also
| 2017 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
| 2017 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
